Chapter 1

1: Given the following host address and subnet mask combinations, determine the subnet address and broadcast addresses:
  • 131.108.1.24 255.255.255.0

  • 151.108.100.67 255.255.255.128

  • 171.199.100.10 255.255.255.224

  • 161.88.40.54 255.255.255.192

A1: Performing a logical AND reveals the following:
  • Subnet 131.18.1.0 and broadcast address 131.108.1.255

  • Subnet 151.108.100.0 and broadcast address 151.108.1.127

  • Subnet 171.199.100.0 and broadcast address 171.199.100.31

  • Subnet 161.88.40.0 and broadcast address 161.88.40.63

2: Given the network 141.56.80.0 and a subnet mask of 255.255.254.0, how many hosts are available on this subnet?
A2: Using the formula 2n-2 = 29-2 = 512 hosts, the subnet mask 255.255.254.0 borrows nine (or n) bits from the subnet mask. ...

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